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First we have to figure out which species is oxidized and which is reduced and balance it in acid solution. On the left side, Ag has an oxidation number of 0 (it is an uncombined element). On the right side, Ag is +1. Why? Ag = +1 + 2CN- = -2 ----------------------------- Ag(CN) ion charge = -1 Note that oxygen on the left side also has an oxidation number of 0 for the same reason as Ag. But on the right side of the equation, it has formed H2O and has a charge of -2. CN- is really a spectator ion (isn't oxidized or reduced) so we can ignore it for now. Let's balance each half-reaction. I'm using = as an arrow sign. Oxidation: Ag = Ag+ + e- (done) Reduction: O2 = H2O Put a 2 in front of H2O O2 = 2H2O to balance oxygen Put 4H+ on the left side O2 + 4H+ = 2H2O to balance hydrogen Put 4e- on the left to O2 + 4H+ + 4e- = 2H2O To add the oxidation and reduction reactions together, I need to multiply the oxidation reaction by 4 so its electrons will cancel with the four that are in the reduction reaction. Doing that you get 4Ag + O2 + 4H+ = 4Ag+ + 2H2O In alkaline solution, you find the H+ above and add an equal number of OH- to BOTH sides of the equation: The 4H+ + 4OH- give 4 H2O. 4Ag + O2 + 4H+ = 4Ag+ + 2H2O +4OH- + 4OH- --------------------------------------... 4H2O If I delete two H2O from each side I get: 4Ag + O2 + 2H2O = 4Ag+ + 4OH- Finally, I can add 2CN- to each side for this: 4Ag + O2 + 2H2O + 8CN- = 4Ag(CN)- + 4OH-

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